A complete and practical solution book for the common school teacher . the note under Fig. 40, DB=| of FB, or 26£ ft. (18) FD=V(FB2-fBT)2)= (19) RO, or DL=LB—BD=22i ft. (20) By similar triangles, FD : DB::R0 : FR, or FR= ft. (21) RD, or OL^DF— FR=+ ft. (22) B0=^(M>+DB*) =+ ft. AOf=V(STLa+T7A2) =+ ft. (23) By similar triangles, DB: FB:: FR: FO, or F0= ft. (24) CO,=V(C7Fa+OrF2)= ft. OE = ^(CkT+CE2) =61 7+ ftf length of the ladder. .-. AO= ft., OB= ft., CO=+ ft., + ft. = the length of the ladder. PROBLEM 354. On the three side
A complete and practical solution book for the common school teacher . the note under Fig. 40, DB=| of FB, or 26£ ft. (18) FD=V(FB2-fBT)2)= (19) RO, or DL=LB—BD=22i ft. (20) By similar triangles, FD : DB::R0 : FR, or FR= ft. (21) RD, or OL^DF— FR=+ ft. (22) B0=^(M>+DB*) =+ ft. AOf=V(STLa+T7A2) =+ ft. (23) By similar triangles, DB: FB:: FR: FO, or F0= ft. (24) CO,=V(C7Fa+OrF2)= ft. OE = ^(CkT+CE2) =61 7+ ftf length of the ladder. .-. AO= ft., OB= ft., CO=+ ft., + ft. = the length of the ladder. PROBLEM 354. On the three sides of any plane triangle construct equilateral*;join the centers of these triangles; prove that the resulting triangle Isequilateral. Solution. (1) Let ABC be any tri- angle, having theequilateral on AB,BC and CA. (2) Let G, H, F be the centers of these tri-angles. (3) Now, we are to prove that GHF is an equi-lateral. Circum-scribe the three tri-angles and GHF bycircles. (4) Draw SPH perpen- dicular, FAC ratherat right angles with. FIG. 47. AC, to intersect AF if produced, in S. 170 FAIRCHIL&S SOLUTION BOOK. (5) AS passes through F; then it bisects /FAC, which makes ZXAC = 30°. (6) In like manner angle QCA=30°, from this it is easy to see that SFP is an equilateral, and angle FPH=120°. (7) .\ FGHP is a quadrilateral and the sum of their opposite angles = 180°, as angle FPH=120°. Angle FGH = 180°—120° = 60°. In the same manner it may be proven thatHFG and FHG are each equal to 60°; therefore FGHis equilateral. Note.—This solution was prepared by the author f or the AmericanMathematical Monthly. PROBLEM 355. A wheel 8 ft. in diameter is run by means of a belt from a wheel 1ft. in radius; the distance between their centers is 12 ft., and the beltis crossed between them, causing- them to revolve in opposite direc-tions: what is the length of the belt used? PROBLEM 356. A ladder 20 ft. long- leans against a perpendicular wall at an an
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